Water coming out from a fountain is modeled by the function f(x) = -x2 + 5x + 4 where f(x) represents the height, in feet, of the water from the fountain at different times x, in seconds. What does the average rate of change of f(x) from x = 3 to x = 5 represent?
rate of change of f(x) between x = a and x = b : \[\large \dfrac{f(b)-f(a)}{b-a}\]
a = 3 b = 5 evaluate f(3), f(5) and plugin
I'm confused @ganeshie8
So (5f - 3f)/5-3 ?
nice try, but no. you need to evaluate : \[\large \dfrac{f(5) - f(3)}{5-3}\]
f(5) is NOT same as 5f f(5) is a notation which represents the function f(x) evaluated at x = 5
you have : f(x) = -x^2 + 5x + 4 can you evaluate f(5) ? f(5) = ?
Ohhh I see now
Give me an minute
f(5) = 54
f(3) = 28
54 -28 / 5 - 3 = 13
careful, its -x^2
But negative times a negative is positive right?
So it wouldn't matter if the negative is there
f(x) = -x^2 + 5x + 4 f(5) = -(5^2) + 5(5) + 4 = -25 + 25 + 4 = 0 + 4 = 4
f(3) = -(3^2) + 5(3) + 4 = -9 + 15 + 4 = 10
the average rate would be : \[\large \dfrac{4 - 10}{5-3}\] simplify
-3
\(\large \color{red}{\checkmark}\)
Notice that by P `E`MDAS, Exponent should be evaluated before Multiplication, when you have an expression like `-x^2`
-x^2 is not same as (-x)^2
So what does the average rate of change of f(x) from x = 3 to x = 5 represent?
@ganeshie8
what do you think it represents ?
\[\large \text{Average rate of change}=\dfrac{f(b)-f(a)}{b-a}\]
That the water comes out of the fountain at a speed of 3 feet per second
numerator has f(x) whose units are `feet` right?
and the denominator has units of seconds `seconds`
so the units of average rate of change would be `feet/sec` which is same as the units of speed, so yeah you're right !
Awesome, thanks
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